The charge, \(q\), on a capacitor in an \(L C R\) series circuit satisfies the
second-order differential equation
$$
L \frac{\mathrm{d}^{2} q}{\mathrm{~d} t^{2}}+R \frac{\mathrm{d} q}{\mathrm{~d}
t}+\frac{1}{C} q=E
$$
where \(L, R, C\) and \(E\) are constants. Show that if \(2 L=C R^{2}\) the general
solution of this equation is
$$
\begin{aligned}
&q= \\
&\mathrm{e}^{-t /(C R)}\left(A \cos \frac{1}{C R} t+B \sin \frac{1}{C R}
t\right)+C E
\end{aligned}
$$ If \(i=\frac{\mathrm{d} q}{\mathrm{~d} t}=0\) and \(q=0\) when \(t=0\) show that
the current in the circuit is
$$
i=\frac{2 E}{R} \mathrm{e}^{-t /(C R)} \sin \frac{1}{C R} t
$$